From 6693aa88adec9df819070e13d7f3ac5aae104365 Mon Sep 17 00:00:00 2001
From: 64f18f326b6383d28a460152fe80edc1
 <64f18f326b6383d28a460152fe80edc1@app-learninglab.inria.fr>
Date: Mon, 6 Sep 2021 12:12:55 +0000
Subject: [PATCH] Replace toy_notebook_en.ipynb

---
 module2/exo1/toy_notebook_en.ipynb | 14 ++++++--------
 1 file changed, 6 insertions(+), 8 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index ead7508..e498ae5 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -4,16 +4,14 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## <center>toy_notebook_en\n",
-    "<center>March 28, 2019"
+    "# 1 On the computation of $\\pi$"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "### **1 On the computation of** $\\pi$\n",
-    "**1.1 Asking the math library**"
+    "## 1.1 Asking the maths library"
    ]
   },
   {
@@ -45,7 +43,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "**1.2 Buffon's needle**"
+    "## 1.2 Buffon's needle"
    ]
   },
   {
@@ -77,21 +75,21 @@
     "N = 10000\n",
     "x = np.random.uniform(size=N, low=0, high=1)\n",
     "theta = np.random.uniform(size=N, low=0, high=pi/2)\n",
-    "2/(sum((x+np.sin(theta))>1)/N)\n"
+    "2/(sum((x+np.sin(theta))>1)/N)"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "**1.3 Using a surface fraction argument**"
+    "## 1.3 Using a surface fraction argument"
    ]
   },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "A method that is easier to understand and does not make use of the sin function is based on the fact that if $X\\sim U(0,1)$ and $Y\\sim U(0,1)$, then $P[X^2+Y^2\\leq 1]=\\pi /4$ (see [\"Monte Carlo method on Wikipedia\"](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
+    "A method that is easier to understand and does not make use of the sin function is based on the fact that if $X\\sim U(0,1)$ and $Y\\sim U(0,1)$, then $P[X^2 + Y^2\\leq 1]=\\pi /4$ (see [\"Monte Carlo method on Wikipedia\"](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:"
    ]
   },
   {
-- 
2.18.1